- #36
collinsmark
Homework Helper
Gold Member
- 3,385
- 2,637
Perfect. Now integrate both sides of the equation. (This will involve two integrals, one on each side of the equation.)vbottas said:
Perfect. Now integrate both sides of the equation. (This will involve two integrals, one on each side of the equation.)vbottas said:
Always try to work in variables till the end. Calculate as last step.vbottas said:should I plug in known values to the left side?
ok! I was just worried taking the integral this many constants labeled as letters may get a bit confusing. But I'll trust you on this!erobz said:Alway try to work in variables till the end.
The constants come outside the integral.vbottas said:ok! I was just worried taking the integral this many constants labeled as letters may get a bit confusing. But I'll trust you on this!
the bounds on the left side is 0 to 45 correct? what would my right side bounds be?collinsmark said:Perfect. Now integrate both sides of the equation. (This will involve two integrals, one on each side of the equation.)
Well, you're integrating over velocity, right? How about [itex] v_f [/itex] and [itex] v_0 [/itex]?vbottas said:the bounds on the left side is 0 to 45 correct? what would my left side bounds be?
One is the speed you began with, the other is variable (final velocity) leave these all variable names though. You can integrate on the left from ##x_o## to ##x_f##, and as was mentioned ##v_o## to ##v_f## on the RHS.vbottas said:the bounds on the left side is 0 to 45 correct? what would my left side bounds be?
i changed the left side back to xi and xf :)vbottas said:
So far so good, except you left out the [itex] A [/itex] by accident somewhere along the line. But yes, that's the right idea.vbottas said:
Keep simplifying. Log rules on the RHS to group those terms, then exponentiate each side.vbottas said:
I can turn the rhs to ln(vf/vi) but that is the only simplificiation I see? Do you agree?erobz said:Keep simplifying
Next. exponentiation of each side.vbottas said:I can turn the rhs to ln(vf/vi) but that is the only simplificiation I see? Do you agree?
Try taking both sides of the equation to the power of e. (I.e., exponentiate each side)vbottas said:I can turn the rhs to ln(vf/vi) but that is the only simplificiation I see? Do you agree?
oh boy its been awhile since i've done this. the rhs ln disappears leaving just vf/vi. but for the left side, does all of the left side become the exponent for e?collinsmark said:Try taking both sides of the equation to the power of e. (I.e., exponentiate each side)
Yes.vbottas said:oh boy its been awhile since i've done this. the rhs ln disappears leaving just vf/vi. but for the left side, does all of the left side become the exponent for e?
'Looks good to me.vbottas said: